Find the equivalent resistance between terminals A and B of the infinite network of resistance as shown in the diagram below :

Explanation

Let the net resistance of this infinite circuit be X. Since the network is infinite, so one more set of resistances `3Omega, 4Omega and 5Omega` will not effect the total resistance X of the network. So equivalent circuit can be drawn as:

Now the `4Omega and X` resistances are in 1 parallel and let their net resistance be `R_p`
So `1/R_P = 1/4 +1/X = (X+4)/(4X)`
`R_P = (4X)/(4+X)`
Now `3Omega, R_p and 5Omega` resistances are in series.
So total resistance across A and B = R
`=3+(4X)/(4+X) +5`
`implies R=8 +(4X)/(4+X)`
But the total resistance of the network remains uneffected.
So R= X
`implies 8+ (4X)/(4+X)=X`
`implies 32 + 8X + 4X= + X^2`
`implies X^2 - 8X - 32 = 0`
On solving, we get `X = 10.92 Omega`